The differential equation of the family of circles,whose centres are on the $X$-axis and which touch the $Y$-axis,is

  • A
    $4(x+y \frac{dy}{dx})^2 x^2 = (x^2+y^2)^2$
  • B
    $(x+y \frac{dy}{dx})^2 x^2 = (x^2+y^2)^2$
  • C
    $2(x+y \frac{dy}{dx})^2 x^2 = (x^2+y^2)^2$
  • D
    $(x+y \frac{dy}{dx})^2 x^2 = 4(x^2+y^2)^2$

Explore More

Similar Questions

The elimination of the arbitrary constants $A, B$ and $C$ from $y = A + Bx + C{e^{ - x}}$ leads to the differential equation:

If the order and degree of the differential equation corresponding to the family of curves $(x-2)^2+(y-a)^2=b^2$,(where $a$ and $b$ are parameters) are $m$ and $n$ respectively,then $m^2+n=$

If $A$ and $B$ are arbitrary constants,then the differential equation having $y=Ae^{x}+B \sin 2 x$ as its general solution is

Verify that the given function $y = \cos x + C$ is a solution of the differential equation $y^{\prime} + \sin x = 0$.

The differential equation of all circles which pass through the origin and whose centre lie on the $Y$-axis is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo